Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+y &= 2 \\ 2x-y &= -3\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $2x = y-3$ Divide both sides by $2$ to isolate $x$ $x = {\dfrac{1}{2}y - \dfrac{3}{2}}$ Substitute this expression for $x$ in the first equation. $-8({\dfrac{1}{2}y - \dfrac{3}{2}}) + y = 2$ $-4y + 12 + y = 2$ Simplify by combining terms, then solve for $y$ $-3y + 12 = 2$ $-3y = -10$ $y = \dfrac{10}{3}$ Substitute $\dfrac{10}{3}$ for $y$ in the top equation. $-8x+ \dfrac{10}{3} = 2$ $-8x+\dfrac{10}{3} = 2$ $-8x = -\dfrac{4}{3}$ $x = \dfrac{1}{6}$ The solution is $\enspace x = \dfrac{1}{6}, \enspace y = \dfrac{10}{3}$.